Last visit was: 24 Apr 2026, 17:46 It is currently 24 Apr 2026, 17:46
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
805+ (Hard)|   Geometry|      
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 24 Apr 2026
Posts: 109,818
Own Kudos:
811,081
 [20]
Given Kudos: 105,873
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,818
Kudos: 811,081
 [20]
In the figure below, rectangles CABD and PMNQ are inscribed in a circl 28 Aug 2015, 02:23
2
Kudos
Add Kudos
18
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

44% (03:47) correct 56% (03:31) wrong based on 55 sessions

History

Date
Time
Result
Not Attempted Yet
In the figure below, rectangles CABD and PMNQ are inscribed in a circle with center O as shown, such that segment CD lies on segment MN, and points A, B, P, and Q lie on the circle. If the radius of the circle is 25, rectangles CABD and PMNQ have the same dimensions, and the shortest distance from O to segment MN is 6, what is the area of the shaded region?




(A) \(625\pi-420\)
(B) \(625\pi-500\)
(C) \(625\pi-640\)
(D) \(625\pi-840\)
(E) \(625\pi-960\)



Kudos for a correct solution.

Show Spoiler
Attachment:
JD_NONSBD_Hard004.GIF
JD_NONSBD_Hard004.GIF [ 4.72 KiB | Viewed 5327 times ]
ShowHide Answer
Official Answer
D
avatar
pluto82
avatar
Joined: 21 Jul 2015
Last visit: 23 Mar 2017
Posts: 27
Own Kudos:
64
 [2]
Given Kudos: 31
Posts: 27
Kudos: 64
 [2]
In the figure below, rectangles CABD and PMNQ are inscribed in a circl 28 Aug 2015, 19:31
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let the width of rectangle NQ = x
Length of rectangle PQ =y
(y/2)^2+(x+6)^2 = 25

Considering Pythagorean triples (3,4,5) => (3*5,4*5,5*5)
y/2 = 15 => y = 30
x+6 = 20 => x = 14

Area or rectangle = 30*14 = 420

Area of shaded region = Area of circle - 2*area of rectangle
= 625*pi - 840

Answer (D)
Attachments

shadedarea.png
shadedarea.png [ 12.28 KiB | Viewed 4784 times ]

avatar
rohitt911
avatar
Joined: 17 Feb 2015
Last visit: 29 Sep 2017
Posts: 14
Own Kudos:
12
 [2]
Given Kudos: 51
Schools: Alberta"19 Degroote McMaster"19 HEC Montreal '20 Ivey '19 Schulich Jan'19 Rotman '19 Desautels '19
WE:Sales (Consulting)
Schools: Alberta"19 Degroote McMaster"19 HEC Montreal '20 Ivey '19 Schulich Jan'19 Rotman '19 Desautels '19
Posts: 14
Kudos: 12
 [2]
In the figure below, rectangles CABD and PMNQ are inscribed in a circl 30 Aug 2015, 00:53
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
pluto82
Let the width of rectangle NQ = x
Length of rectangle PQ =y
(y/2)^2+(x+6)^2 = 25

Considering Pythagorean triples (3,4,5) => (3*5,4*5,5*5)
y/2 = 15 => y = 30
x+6 = 20 => x = 14

Area or rectangle = 30*14 = 420

Area of shaded region = Area of circle - 2*area of rectangle
= 625*pi - 840

Answer (D)
Show more


To determine the area of shaded region : {(Area of circle)- (Area of ABDC +Area of MNQP)}

We are given the radius (25 units).

Let R be mid point of MN such that OR =6 units (the shortest distance from O to segment MN is 6)

Let T be a point on PQ such that RT = x units

OT =OR + RT = 6+x units

Let TQ = y units.

OTQ is a right angled triangled at T

Using Pythagorean Principle :

OT ^2 +TQ ^2 = OQ
(6+x)^2 + y ^2 = 25 ^2

The triplets could be (15, 20) or (20 ,15). So length & breadth of the rectangles can be (9, 20) or (14,15).

Case 1:(9, 20*2)
Area of the rectangle : 360 units
Area of 2 rectangle : 2 * 360 : 720 units

Area of shaded region = area of circle - area of 2 rectangles.
========= 625 pie - 720

Case 2:(14 , 15*2)
Area of rectangle : 420 units
Area of 2 rectangles : 840 units

Area of shaded region :area of circle - area of 2 rectangles.
= 625 pie - 840.

Case 2 matches with option D
User avatar
sahilvijay
User avatar
Joined: 29 Jun 2017
Last visit: 16 Apr 2021
Posts: 289
Own Kudos:
931
Given Kudos: 76
GPA: 4
WE:Engineering (Transportation)
Products:
My Reviews
Posts: 289
Kudos: 931
In the figure below, rectangles CABD and PMNQ are inscribed in a circl 04 Sep 2017, 07:46
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Ans is D:

HERE I HAVE SOLVED BY ASSUMPTION METHOD ASSUMING THAT THEY ARE FORMING TRIPLET TAKING INTO ACCOUNT THAT L AND B ARE INTEGERS

I tried solving by other methods also but could not solve.

Although by assumption i got 2 values of shaded area of which one is matching my answer, that is option D .

Apart from that we could have once more possible answer , Luckily that is not available
i have shown both answers in my image.
Attachments

IMG_6886.JPG
IMG_6886.JPG [ 1.39 MiB | Viewed 3754 times ]

User avatar
firas92
User avatar
Current Student
Joined: 16 Jan 2019
Last visit: 02 Dec 2024
Posts: 616
Own Kudos:
1,766
Given Kudos: 142
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
Products:
My Reviews
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
Posts: 616
Kudos: 1,766
In the figure below, rectangles CABD and PMNQ are inscribed in a circl 18 Aug 2019, 06:25
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let the length of the rectangle be \(x\) and breadth be \(y\).

From the attached diagram, we can draw two equations

\((x-6)^2+(\frac{y}{2})^2=25^2\) ---------------- (I)
\((\frac{x}{2})^2+(y+6)^2=25^2\)---------------- (II)

Equating (I) and (II)

\(x^2-12x+36+\frac{y^2}{4}=\frac{x^2}{4}+y^2+12y+36\)
\(x^2-\frac{x^2}{4}+\frac{y^2}{4}-y^2=12x+12y\)

\(\frac{3x^2}{4}-\frac{3y^2}{4}=12(x+y)\)
\(\frac{3(x^2-y^2)}{4}=12(x+y)\)

\(\frac{(x+y)(x-y)}{4}=4(x+y)\)
\(x-y=16\) and so

\(x=y+16\)

Substitute for \(x\) in (I)

\((y+16-6)^2+(\frac{y}{2})^2=25^2\)
\((y+10)^2+\frac{y^2}{4}=625\)

\(y^2+20y+100+\frac{y^2}{4}=625\)
\(\frac{5y^2}{4}+20y-525=0\)

\(\frac{y^2}{4}+4y-105=0\)
\(y^2+16y-420=0\)

\((y+30)(y-14)=0\)

Therefore, \(y=14\) and so \(x=16+14=30\)

Area of rectangle is \(14*30=420\)
So combined area of the two rectangles is \(420*2=840\)

Therefore area of the shaded region is \(625\pi-840\)

Answer is (D)
Attachments

Untitled.png
Untitled.png [ 31.94 KiB | Viewed 2959 times ]

User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,975
Own Kudos:
1,117
Posts: 38,975
Kudos: 1,117
In the figure below, rectangles CABD and PMNQ are inscribed in a circl 29 Oct 2020, 04:00
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Bunuel
Math Expert
109818 posts
Krunaal
Tuck School Moderator
853 posts